Functions

Definition

Let FA×BF\subseteq A\times B so its a relation

Suppose F has the property that if aAa\in A appears as a a first entry in an element of FF, then it is the only element of FF that it appears in

Then we call FF a function

If (a,b)F(a,b)\in F, we write f(a)=bf(a) = b


In a function, second entries can be matched with more than one first entry

If, FA×BF\subseteq A\times B is a function with f(a)=b iff (a,b)Ff(a)=b\ iff\ (a,b) \leftarrow F we often omit FF and only refer to ff and say:

f:ABf:A\rightarrow B
bdg-infoWe say ff maps AA into BB

Consider the following example

A &= \Set{0,1,2,3,4,5,6,7,8,9}\\ B &= \Set{r,s,t,u,v,w,x,y,z}\\ F &= \Set{(0,r), (1,s), (2,s), (3,t), (4,u), (5,w), (6,x), (7,x), (8,x), (9,x) }

Then the rule for ff may be stated explicitly or inferred from FF

bdg-warningNote range(f)=\Setr,s,t,u,v,w,x,y,zBrange(f) = \Set{r,s,t,u,v,w,x,y,z} \subset B is a proper subset of BB

Let us write f1(b)=f^{-1}(b)= pre-image of bb under ff =\SetaAb=f(a)=\Set{a\in A\mid b = f(a)}

For our example

f1(z)=f^{-1}(z)=\emptyset
f1(t)=\Set3, f1(x)=\Set6,7,8,9f^{-1}(t)=\Set{3},\ f^{-1}(x)=\Set{6,7,8,9}

Composition of function

Let f:ABf: A \rightarrow B & g:BCg: B \rightarrow C be two function

We define the composition of gg and ff to be the function gf:ACg\circ f:A\rightarrow C defined by gf(a)=g(f(a))g\circ f(a)=g(f(a))

Composition of More Than 2 Function

We can define the composition of more than 2 functions if f:ABf: A \rightarrow B, g:BGg: B \rightarrow G, h:GDh: G \rightarrow D, then

hgf:ADh\circ g\circ f:A\rightarrow D and hgf(a)=h(g(f(a)))h\circ g\circ f(a)=h(g(f(a))) etc

Inverse Functions

Let f:ABf:A\rightarrow B be a 1-1 onto map

Then for bB, exactly one aA st f(a)=b\forall b\in B, \exists\text{ exactly one }a\in A\ st\ f(a)=b

We define the inverse function of ff, called f1(b)=af^{-1}(b)=a iff f(a)=bf(a)=b and we say ff is invertible

Identity Function

If we let idA:AAid_A:A\rightarrow A be the identity map on AA, idA(a)=a,aAid_A(a)=a,\forall a\leftarrow A and idB:BBid_B:B\rightarrow B be the identity map on B, idB(b)=bB,\ id_B(b)=b, then for ff, 1-1, onto, f1f=idAf^{-1}\circ f =id_A, ff1=idBf\circ f^{-1}=id_B